lim-x-sinx-x-3-2sinx-

Question Number 158175 by mathocean1 last updated on 31/Oct/21

\underset{x \to + \infty}{l i m} \frac{s i n x + x}{3 + 2 s i n x} = ?

Commented by cortano last updated on 01/Nov/21

\lim_{x \to \infty} \frac{\sin x + x}{3 + 2 \sin x} = \lim_{x \to \infty} \frac{\frac{\sin x}{x} + 1}{\frac{3}{x} + \frac{2 \sin x}{x}}

= \frac{0 + 1}{0 + 0} = \infty

Answered by puissant last updated on 31/Oct/21

u = \frac{1}{x}; u \to 0 q u a n d x \to + \infty

\Rightarrow L = \lim_{x \to + \infty} \frac{s i n x + x}{3 + 2 s i n x} = \lim_{u \to 0} \frac{s i n (\frac{1}{u}) + \frac{1}{u}}{3 + 2 s i n (\frac{1}{u})}

= \lim_{u \to 0} \frac{\frac{1}{u} + \frac{1}{u}}{3 + \frac{2}{u}} (e n u t i l i s a n t l e d e v e l l o p e m e n t

l i m i t \overset{´}{e} d u s i n u s a l^{'} o r d r e 1 . .

\Rightarrow L = \lim_{u \to 0} \frac{\frac{2}{u}}{\frac{3 u + 2}{u}} = \lim_{u \to 0} \frac{2}{3 u + 2} = 1

Facebook Tweet Pin